So how do you find those chains (xy-chain)? Do you start at every pair until you find one (or dont), or is there some way of spotting them?

I began to find them as sort of an extension of an XY-wing. In this case, notice that you can collapse 36 16 14 34 in a number of ways. Suppose you are looking for an XY-wing, and notice 36 16. So, you are looking for 13 to complete the wing. Together, 14 and 34 behave like a "pseudo-cell" 13.

So, if I can't find any XY-wings, I look at pairs of cells like WZ and WY. Together they make XZ, can I fit that in a pattern?

Or, if an XY-wing is a 3-cell chain

XZ - XY - YZ, with pincers Z,

the 4-cell chain is

XZ - XY - WY - WZ, with pincers Z.

It took me some practice, a couple of weeks, but I now find these not much harder to spot than XY-wings.

First, noting that "r7c6=4" should really be "r7c6=3", you could call it an "XYZ-Wing with Transport". The XYZ-Wing is on 347 in r3c69 and r7c6. By itself, this is useless because of the geometry: there are no common peers of those three cells so no <4> eliminations. However, the two <4>s in r3c69 (considered together as a group) can be transported to r9c3 (via r3c3). Then, r7c6 and r3c3 work together as pincers to eliminate <4> from r9c6.

You could work it the other way and transport r7c6 to r9c3 (via r9c6) and eliminate <4> from r3c3. The <4> at r9c3 and the grouped <4>s at r3c69 are the pincers in that case.

Sometimes what look like useless wing patterns can be useful if a pincer can be transported.

Asellus, you have again exposed me to another solution technique; thanks!

Looking at the posted code, I believe I found still another view of the xyz-wing with transport. The xyz-wing is r39c6 and pseudo-cell <34> in r3c13. Then transport <4> in r9c6 to r3c3 via r9c3 to delete <4> in r3c9.

Looking at the posted code, I believe I found still another view of the xyz-wing with transport. The xyz-wing is r39c6 and pseudo-cell <34> in r3c13. Then transport <4> in r9c6 to r3c3 via r9c3 to delete <4> in r3c9.

It is called Eureka notation. It says:
If you assume r3c6=7,
then r9c6 <>7, it is 4,
then r9c3 <> 4
then r3c3 = 4,
then r3c9 <> 4, it is 7.
This would create an invalid condition since two <7s> would be in row 3; so r3c6 <>7.

And yes, gindaani was using forcing (SIN, etc.) in the original description of the <4> elimination. I just wanted to show that resort to forcing wasn't necessary.

As for Eureka notation, the "if this is true, then that is false, etc." approach is one way to understand it. Better, in my opinion, is to learn to see it in terms of the inferences and Alternate Implication Chains (AIC). The terms and concepts involved are explained in various places online (such as Sudopedia). [Note: I hasten to add that someone still getting started in advanced sudoku solving techniques should probably focus most of all on learning the common "wings" and other techniques, and then ease into learning about these implications, etc.]

Looking at the blue part of the notation, the <7> and <4> in r9c6 have a strong inference (cannot both be false), denoted by the "=" symbol, because they form a bivalue cell. The two <4>s in r9c3 and r3c3 have a strong inference because they are the only two <4>s in column 3. And again, in r3c9, we have a bivalue cell for another strong inference.

The <4>s in r9c6 and r9c3 have a weak inference (cannot both be true), denoted by the "-" symbol, because they share a house, row 9 in this case. Similarly, the <4>s in r3c3 and r3c9 have a weak inference in row 3.

The inferences ("implications") in the blue section alternate strong-weak-strong-weak-strong, or "= - = - =". In such a chain of alternating implications (AIC), the implications propagate. So, the <7>s that are connected to the two ends, at r9c6 and r3c9, by strong inferences themselves have a strong inference between them. That is, they cannot both be false. Any <7> that can "see" both of these <7>s (or more generally, any <7>s with a weak inference to both of these) cannot be true. The (non-blue) <7> at r3c6 shown attached by a weak inference on both ends, thus cannot be true. This conclusion is stated explicitly after the semicolon. (Some people use "=>" in place of the semicolon. But I don't like having a "=" sign in the notation that does not mark a strong inference so prefer the semicolon.)

It's not as easy to describe as the "If... then..." approach! But, it is ultimately a more revealing and useful way to see the notation. This description is not complete: several additional things could be mentioned. But, it paints an adequate introductory picture, I believe.